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INCM 9102
Quantitative Methods
ANOVA
ANOVA
Previously, we covered several different forms of t-tests.
For example, we used ttests to answer questions such as:
Are cars going over the speed limit on this residential street?
Did students who took a preparatory course, score better on the
standardized exam than students who did not take the course?
ANOVA
The general setting for all of these questions is that some
quantitative variable (speed, test scores) has been measured for
one or two categories of subjects.
What if we have more than two categories across which we want
to compare the value of some quantitative variable?
For example, lets say that we wanted to compare the mean weight
loss of subjects who were put on one of four diet plans. For ease
of discussion, lets call these plans A, B, C and D.
ANOVA
The following approach would be tempting…
H0: Plan A = Plan B
H0: Plan B = Plan C
H1: plan A  Plan B
H1: plan B  Plan C
H0: Plan C = Plan D
H0: Plan A = Plan C
H1: plan C  Plan D
H1: plan A  Plan C
H0: Plan A = Plan D
H0: Plan B = Plan D
H1: plan A  Plan D
H1: plan B  Plan D
ANOVA
…but wrong.
Apart from being very cumbersome, there is a critical problem –
we are inflating our probability of making a type 1 error.
Think about that – lets use alpha = .05. If we ran 6 separate tests,
that would generate a cumulative probability of a type 1 error of
.3.
We could lower the alpha value to .05/6 – I hear you saying. But
this has its own problems – what happens if the number of tests
increase to 8 or 10? Our alpha value would become so low, we
would almost never reject the null (recall Power).
ANOVA
What we need is a single test which will allow us to evaluate all
of the relationships simultaneously while using a reasonable
alpha level.
The test needs to be able to provide information regarding
differences in the mean values of multiple subject groupings.
To accomplish this, we use the Analysis of Variance or ANOVA
procedure.
ANOVA
Lets discuss how to use ANOVA to test a hypothesis by returning
to our dieters…
In this instance there are four levels (diet plans) to a single factor
(weight loss).
The hypothesis statements would look like this:
H0: All level means are equal. In other words, all four of the diet
plans generate approximately the same amount of weight loss.
H1: Not all of the level means are equal. In other words, at least one
of the plans’ weight loss mean is statistically significant different from
the other plans’ means.
ANOVA
Prior to executing the test, we must check for three important
assumptions about our data:
1.
2.
3.
All the groups are normally distributed.
All the populations sampled have approximately equal
variance (you can check this by generating side-by-side
boxplots). The rule of thumb is that the largest std is <2x the
smallest std.
The samples of the groups are independent of each other and
subjects within the groups were randomly selected.
As with most, but not all, statistical tests, if our samples are large,
we can relax our assumptions and work around non normal data.
ANOVA
Lets examine the hypothesis statements in more detail:
H0: µa = µb = µc = µd
H1: µa ≠ µb ≠ µc ≠ µd
Consider – what would the hypothesized distributions look like
under H0 and H1?
ANOVA
Ok. We understand the concept, we have the hypotheses, we
have the assumptions – we need a test statistic.
In ANOVA, we use the F-distribution. In the science of statistics,
whenever you need to evaluate a ratio of variances you will be
using an F-statistic.
The ratio in question here is:
The variation BETWEEN the groups
The variation WITHIN the groups
Question – what kind of value would indicate difference versus
no difference?
ANOVA
The result of this ratio is the F-statistic. You can see the FDistribution in your book. As the number of groups and
observations increases, the distribution will start to appear
normal.
Lets start working with an example…
ANOVA
Returning to the diet plans…
PLAN Mean
PLAN A
14
14
20
22
26
27
20.50
PLAN B
15
18
23
25
28
30
23.17
PLAN C
32
36
40
42
45
45
40.00
PLAN D
33
38
42
44
46
47
41.67
OVERALL MEAN
31.33
ANOVA
Our hypotheses statements would be:
H0: The four diets plans have the same results (the mean weight
loss is the same)
H1: At least one of the diet plans has a different result (the mean
weight loss is different)
We will now calculate our test statistic:
The variation BETWEEN the groups
The variation WITHIN the groups
ANOVA
To calculate the F-Statistic, we use the following table (refer to page 682
in your book):
SOURCE
SUM OF
SQUARE
S
DEGREES
OF
FREEDOM
MEAN
SQUARE
F-stat
BETWEEN
SSB
# levels – 1
SSB/(# levels – 1)
{SSB/(# levels – 1)}
{SSW(n- # levels)}
WITHIN
SSW
n- # levels
SSW/(n- # levels)
TOTAL
SST
(SSB + SSW)
n-1
ANOVA
For those who are interested:
SST
_
= SSW
_
+ SSB
_
ij(Xij-X)2 = ij(Xij-Xj)2 + nj(Xj-X)2
ANOVA
For the present problem:
SOURCE
SUM OF
SQUARE
S
DEGREES
OF
FREEDOM
MEAN
SQUARE
BETWEEN1
2195.67
3
731.89
WITHIN2
601.67
20
30.08
TOTAL
2797.34
23
1
SSB = 6(10.832 + 8.172 + 8.672 +10.332)
2SSW
= (159.50 + 166.83 + 134 + 141.33) = 601.67
F-stat
24.33
ANOVA
Now…what to do with an F-statistic of 24.33?
This is a fairly strong statistic – recall that as the variance ratio
approaches 1, the null is considered to be true. As the variance ratio
grows larger than 1, we can more confidently reject the null.
As with all test statistics, this result will translate into a p-value. The pvalue associated with this statistic is less than .001. Based upon this
result, we can confidently reject the null hypothesis and conclude that at
least one of the results is different.
Lets execute this same problem in SPSS…